Xiao Ming and Xiao Liang set off from A and B at the same time and traveled in the opposite directio

Xiao Ming and Xiao Liang set off from A and B at the same time, Xiao Ming's speed was five-sixths of Xiao Liang's, and after the two arrived at B and A respectively, they immediately returned to their respective starting places. The speed at the time of returning, Xiao Ming increased by one-fifth compared with the original, Xiao Liang increased by one-fourth compared with the original, and it is known that the first meeting between the two was 35 kilometers away from the second meeting on the way back. How many kilometers are between A and B?

1) If the starting speed of Xiaoliang is V, then Xiaoming's speed is 5V 6, and when he returns, Xiaoming's speed is V, and Xiaoliang's speed is 5V 4

Let the distance between the two places be s

The distance from the first place is 5s 11

The distance from the second place is 2s 3

So 35 = 2s 3 - 5s 11 s = 1652) A and B are 165 kilometers apart.

Let the total distance of A and B s, at the beginning Xiao Liang's velocity is v, and Xiao Ming's velocity is (5 6) v. The speed of Xiao Ming after turning is (5 6)v*(1+1 5)=v, and the speed of Xiao Liang after turning is v*(1+1 4) = (5 4)v, and the initial moment is 0

The time of their first encounter is s (v + (5 6)v) = (6 11)*(s v).

The place where the two first met was from place B.

6 11)*(s v)*v = (6 11)s The time for Xiaoliang to arrive at A is sv

The time for Xiao Ming to arrive at place B is ((5 6)v) = (6 5)(s v), so the time before Xiao Ming arrives at place B after Xiao Liang arrives at place A is.

6 5)(s v)-s v = (1 5)(s v) During this time, Xiaoliang walked a distance of .

1 5)(s v)*(5 4)v = (1 4)s, that is, after Xiao Ming arrives at place B, the distance between the two is.

s-(1/4)s = (3/4)s

The place where the two meet for the second time is from place B.

3 4)s (v+(5 4)v)] v = (1 3)s The distance between the two meeting points is .

6 11)s - 1 3)s = (7 33)s = 35 km.

s = 35 (7 33) = 165 km.

I have been reading this question for a long time, and the method given is not suitable for understanding, so I made some on paper myself and the majority of students can understand.

Let the speed of Xiao Ming and Xiao Liang be x km h and y km h (x + y) 2=20 respectively

2x-2y=2 (because it took 2 hours for Xiao Ming to return to A) to solve the equation system to get x=,y=

Answer: Xiao Ming's speed is kilometers per hour, and Xiao Liang's speed is kilometers per hour.

When the speed of Xiao Ming is x km, the speed of Xiao Liang is 2(x+y)=20 when it is y km

2（x-y）=2

Solution: x= y=

A: ......

Let the distance between A and B be x meters.

met for the first time, Xiao Ming walked 680 meters, and the two walked x meters together. The second time they met, the two walked 3x meters together, then Xiao Ming walked 680x3=2040 meters, and the distance between A and B was 2040-340=1700 meters.

The first full journey Xiao Ming walked a 680, from the beginning to the second meeting they walked a total of three full courses, then Xiao Ming walked a total of 3 680, which includes Xiao Ming from B back to 340, so A and B are far apart: 680 * 3 - 340 = 1700

Xiao Ming walked for a minute and walked 60 meters.

It's these 60 meters that I'm chasing.

60 (75-60) = 4 points.

Xiao Hua walked for 4 minutes to catch up with Xiao Ming.

The average speed of the two is x kilometers per hour, y = 5 (x-1), and the distance between the two places is y kilometers. If the solution is x=5, then y=4x

Let Xiao Liang's starting speed be V, then Xiao Ming's speed is 5v 6, and Xiao Ming's speed when he returns is V, Xiao Liang's speed is 5v 4, and the distance between the two places is S

The distance from the first place is 5s 11

The distance from the second place is 2s 3

So 35 = 2s 3 - 5s 11 s = 165 A and B are 165 kilometers apart.